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In his wonderfully titled paper Can you take Solovay's inaccessible away? Shelah showed that if every $\mathbf{\Sigma}^1_3$ set of reals is Lebesgue measurable, then $\omega_1$ is an inaccessible cardinal in $L$. More specifically he shows that if there is a real $a$ such that $\omega_1^{L[a]} = \omega_1$, then there is a non-measurable $\mathbf{\Sigma}^1_3$ set of reals.

Question. Assume that $\omega_1$ is not inaccessible in $L$. Is there some kind of bound on the Turing degrees of reals $a$ such that there is a non-measurable $\Sigma^1_3[a]$ set of reals?

In a remark Shelah mentions that the proof only really uses two parameters, a $b$ such that $\omega_1^{L[b]} = \omega_1$ and a $c$ such that $\bigcup\{B : B~\text{a Borel set of measure zero which has a code in}~L[c]\}$ is non-measurable.

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  • $\begingroup$ Is it clear that $a=\emptyset$ doesn't work (e.g. after Sacks forcing once over $L$)? $\endgroup$
    – Noah Schweber
    Commented Nov 20 at 3:03
  • $\begingroup$ @NoahSchweber I don't know but I'm asking for a bound that works in any model satisfying that $\omega_1$ is not inaccessible in $L$. Does this argument let you establish something like this? $\endgroup$
    – James E Hanson
    Commented Nov 20 at 3:06
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    $\begingroup$ In $L^{\mathrm{Col}(\omega, \omega_1)}$ all OD sets of reals are measurable, so I don't think there's going to be any reasonable answer to this question. $\endgroup$
    – Elliot Glazer
    Commented Nov 20 at 5:35
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    $\begingroup$ Reference for the above claim^ irif.fr/~krivine/articles/Modeles_de_ZF_CRAS.pdf $\endgroup$
    – Elliot Glazer
    Commented Nov 20 at 6:01
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    $\begingroup$ I thought you were asking about both upper and lower bounds. My comment was towards the idea that there may be no always-applicable lower bound. $\endgroup$
    – Noah Schweber
    Commented Nov 20 at 20:13

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